Problem description:

Given a binary tree, return the zigzag level order traversal of its nodes’ values. (ie, from left to right, then right to left for the next level and alternate between).

For example:
Given binary tree [3,9,20,null,null,15,7],

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  3
 / \
9  20
  /  \
 15   7

return its zigzag level order traversal as:

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[
  [3],
  [20,9],
  [15,7]
]

Solution1 BFS+ Deque:

Use a deque to store the elements. The idea is to change direction of push elements into the deque and pop elements from deque.

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def zigzagLevelOrder(self, root: TreeNode) -> List[List[int]]:
        if not root:
            return []
        q, res, d = deque([]), [], True
        # True: look from right to left
        # False: look from left to right
        q.append(root)
        while q:
            pre_level_size = len(q)
            level = []
            for _ in range(pre_level_size):
                if d: 
                    # cur level pop from right, append next level from left
                    # pop from right so we would the rightest node as first one
                    tmp = q.pop()
                    if tmp.left: q.appendleft(tmp.left)
                    if tmp.right: q.appendleft(tmp.right)
                else: 
                    # cur level pop from left, append next level from right
                    # pop from left so we would the leftest node as first one
                    tmp = q.popleft()
                    if tmp.right: q.append(tmp.right)
                    if tmp.left: q.append(tmp.left)
                level.append(tmp.val)
            d = not d
            res.append(level)
        return res
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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        vector<vector<int>> res;
        if(!root) return res;
        deque<TreeNode*> deq;
        bool dir= false;
        deq.push_front(root);
        while(!deq.empty()){
            int size= deq.size();
            vector<int> curlevel;
            for(int i= 0; i< size; i++){
                if(dir){
                    auto tmp= deq.front(); deq.pop_front();
                    curlevel.push_back(tmp->val);
                    if(tmp->right) deq.push_back(tmp->right);
                    if(tmp->left) deq.push_back(tmp->left);
                }
                else{
                    auto tmp= deq.back(); deq.pop_back();
                    curlevel.push_back(tmp->val);
                    if(tmp->left) deq.push_front(tmp->left);
                    if(tmp->right) deq.push_front(tmp->right);
                }
            }
            dir = dir^1;
            res.push_back(curlevel);
        }
        return res;
    }
};

time complexity: $O(n)$
space complexity: $O(n)$

Solution2 BFS+ two stacks:

Use two stacks, one for odd layer, the other one for even layer.

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        vector<vector<int>> res;
        if(!root) return res;
        stack<TreeNode*> odd, even;
        odd.push(root);
        
        while(!odd.empty() || !even.empty()){
            vector<int> level_odd;
            while(!odd.empty()){
                auto tmp= odd.top(); odd.pop();
                if(tmp->left) even.push(tmp->left);
                if(tmp->right) even.push(tmp->right);
                level_odd.push_back(tmp->val);
            }
            if(!level_odd.empty()) res.push_back(level_odd);
            vector<int> level_even;
            while(!even.empty()){
                auto tmp= even.top(); even.pop();
                if(tmp->right) odd.push(tmp->right);
                if(tmp->left) odd.push(tmp->left);
                level_even.push_back(tmp->val);
            }
            if(!level_even.empty()) res.push_back(level_even);
        }
        return res;
    }
};

time complexity: $O(n)$
space complexity: $O(n)$
reference: